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Explicit and recursive formulas, arithmetic and geometric terms and sums, sigma notation, Pascal's triangle and binomial expansions.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson patterns become formulas. An arithmetic sequence adds a fixed amount and a geometric one multiplies by a fixed factor, and each gets a formula for its $n$th term that saves you from listing the ones before it; adding a whole arithmetic sequence turns out to be one multiplication once you notice that the ends pair up. You also learn sigma notation, which packs a long sum into one symbol, and the difference between an explicit rule and a recursive one that has to be walked. Then the binomial theorem gives every coefficient of $(a+b)^n$ at once — and the same numbers turn out to count choices.
You can continue a pattern like $3, 7, 11, \ldots$ and you can substitute a number into a formula. You have probably also met the trick of adding $1 + 2 + \cdots + 100$ by pairing the ends. This lesson makes both of those into formulas you can apply without re-deriving them, and introduces the notation mathematics uses to write a long sum in one symbol.
Sequence: an ordered list of numbers; $a_n$ is the $n$th one.
Series: the sum of a sequence's terms.
Arithmetic: each term is the previous one plus a fixed common difference $d$.
Geometric: each term is the previous one times a fixed common ratio $r$.
Explicit formula: gives $a_n$ straight from $n$.
Recursive formula: gives $a_n$ from $a_{n-1}$, plus a starting value.
Sigma notation: $\sum_{k=1}^{n}$ means add, for $k = 1$ up to $n$.
An arithmetic sequence adds a fixed amount each time, so $$a_n = a_1 + (n-1)d,$$ where the $n - 1$ is the number of steps taken, not the number of terms reached: $7, 11, 15, \ldots$ has $a_{20} = 7 + 19 \times 4 = 83$. Its sum comes from pairing the ends — first with last, second with second-last, every pair the same total — giving $$S_n = \frac{n}{2}\left(a_1 + a_n\right),$$ which for $1 + 2 + \cdots + 100$ is $50 \times 101 = 5050$. A geometric sequence multiplies instead: $a_n = a_1r^{\,n-1}$, so $3, 6, 12, \ldots$ has $a_8 = 3 \times 2^7 = 384$. Sigma notation packs a sum into one symbol: $\sum_{k=1}^{5}(2k+1)$ means $3 + 5 + 7 + 9 + 11 = 35$, and it splits, $\sum(ak + b) = a\sum k + nb$. Finally, a recursive rule gives each term from the one before it and must be walked one step at a time: $a_1 = 2$, $a_n = 3a_{n-1} - 1$ gives $2, 5, 14, 41, \ldots$.
Another way: picture
The sum $3 + 5 + 7 + 9$ drawn as four columns of dots of increasing height, with a second upside-down copy fitted on top: the two together make a perfect rectangle $4$ wide and $12$ tall, so the original is half of $4 \times 12$.
Another way: steps
Multiplying by $n$ instead of $n - 1$. The $20$th term of $7, 11, 15, \ldots$ is $7 + 19 \times 4 = 83$, not $7 + 20 \times 4$. The first term has taken no steps at all.
"Sum $= n \times$ last term." That would count every term as the biggest one. The right factor is the average of the first and last, so Sum $= \frac{n}{2}(a_1 + a_n)$.
Trying to jump ahead in a recursive rule. With $a_1 = 2$ and $a_n = 3a_{n-1} - 1$ there is no way to get $a_5$ without $a_4$. Recursive rules are walked, not evaluated.
It is arithmetic with $a_1 = 4$ and $d = 5$, so $a_n = 4 + 5(n-1)$.
Write the explicit formula first.
Set $4 + 5(n-1) = 99$: $5(n-1) = 95$, so $n - 1 = 19$ and $n = 20$.
Solve for the position, not the value.
Split it: $2\sum_{k=1}^{5}k + \sum_{k=1}^{5}1 = 2(1+2+3+4+5) + 5 \times 1$.
The sigma distributes over the sum.
$2 \times 15 + 5 = 35$, which matches $3+5+7+9+11$.
$d = 3$, so the tenth term is $2 + 9 \times 3 = 29$.
Sum $= \frac{10}{2}(2 + 29) = 5 \times 31 = 155$.
An arithmetic sequence starts at $11$ and goes up by $8$ each time. What is its $18$th term?
The nth term is answer.
Add the first $20$ terms of the arithmetic sequence starting at $8$ with common difference $5$.
The sum is answer.
A geometric sequence starts at $5$ and multiplies by $2$ each time. What is its $7$th term?
The nth term is answer.
Evaluate $\displaystyle\sum_{k=1}^{8} \left(k + 2\right)$.
The sigma sum is answer.
You can expand $(x + 2)^2$ by multiplying out, and you have probably built a few rows of Pascal's triangle by adding pairs. You also know that exponents count repeated multiplication. The binomial theorem is what saves you from multiplying out $(x + 2)^7$ by hand — and, less obviously, it is a counting statement.
Binomial: a two-term expression such as $a + b$.
Binomial coefficient $\binom{n}{k}$, read '$n$ choose $k$': the number of ways to choose $k$ things from $n$, and the entry in row $n$, position $k$ of Pascal's triangle.
Factorial $n!$: $n \times (n-1) \times \cdots \times 1$.
Pascal's triangle: rows built by adding neighbouring pairs, starting from a single $1$.
Expansion: the result of multiplying everything out and collecting terms.
The binomial theorem says $$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{\,n-k} b^{\,k},$$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. So $(x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1$, using row $4$ of Pascal's triangle, $1\;4\;6\;4\;1$. Three facts fall straight out. There are $n + 1$ terms, one for each power of $a$ from $n$ down to $0$. The row is symmetric, $\binom{n}{k} = \binom{n}{n-k}$, because choosing what to take is choosing what to leave. And the row adds to $2^n$, because putting $a = b = 1$ makes the left side $2^n$. Two cautions in use: the constant carries a power too, so the $x^2$ term of $(x+2)^5$ is $\binom{5}{2}x^2 2^3 = 80x^2$; and the coefficient you want is rarely the one on the end. The second reading is the important one: $\binom{n}{k}$ counts. It is the number of ways of choosing $k$ things from $n$, which is why it appears when you expand — each term counts the ways of picking $b$ from $k$ of the brackets.
| $n$ | row of $\binom{n}{k}$ | row total |
|---|---|---|
| $0$ | $1$ | $1$ |
| $1$ | $1$ $1$ | $2$ |
| $2$ | $1$ $2$ $1$ | $4$ |
| $3$ | $1$ $3$ $3$ $1$ | $8$ |
| $4$ | $1$ $4$ $6$ $4$ $1$ | $16$ |
| $5$ | $1$ $5$ $10$ $10$ $5$ $1$ | $32$ |
Another way: picture
Pascal's triangle to row six, with two adjacent entries of one row circled and an arrow running down to the entry beneath them that they add to, and the row totals $1, 2, 4, 8, 16, 32, 64$ written down the right-hand side.
Another way: steps
"$(x + 2)^5$ has $x^5 + 2^5$ in it and nothing else." It has six terms. A power of a sum is not the sum of the powers, and the middle terms are where all the size is.
Forgetting the powers of the constant. In $(x + 2)^5$ the coefficient of $x^2$ is $\binom{5}{2} \times 2^3 = 10 \times 8 = 80$, not $10$. Each term carries a power of the second thing too.
"Row $6$ has six entries." It has seven — the rows are numbered from zero, and row $n$ has $n + 1$ entries, matching the $n+1$ terms of the expansion.
The general term is $\binom{5}{k}x^{k}2^{5-k}$; for $x^2$ take $k = 2$.
Match the power you want, then read off the other.
$\binom{5}{2} = 10$ and $2^{3} = 8$, so the coefficient is $10 \times 8 = 80$.
The constant carries a power too.
Row $6$ is the list of coefficients in $(a + b)^6$.
The row and the expansion are the same object.
Put $a = b = 1$: the left side is $2^6 = 64$, and the right side is $1+6+15+20+15+6+1$, the whole row.
Powers of $a$ run from $10$ down to $0$, so there are $11$ terms.
The middle entry of row $6$ is $\binom{6}{3} = 20$.
Evaluate $\dbinom{6}{2}$.
The binomial coefficient is answer.
What is the coefficient of $x^{1}$ in the expansion of $(x + 2)^{5}$?
The coefficient is answer.
What do the entries of row $7$ of Pascal's triangle add up to?
The row sum is answer.
In the expansion of $(a + b)^{5}$, how many terms are there? And $\dbinom{5}{3}$ is equal to $\dbinom{5}{j}$ for which other $j$?
terms = t, j = j
A coin is tossed $7$ times and the result is written down as a string like HTTHT. In how many of those strings do exactly $5$ of the tosses come up heads?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A sequence has $a_1 = 4$ and $a_n = 3a_{n-1} - 3$. Give $a_2$ and then $a_3$.
a2 = p, a3 = q
Expand $(x + 1)^{3}$ fully. Write the answer with no brackets.
Answer:
You can find terms and sums of sequences and expand a binomial. Without looking: why is the $20$th term of $7, 11, 15, \ldots$ found with $19$ steps and not $20$, and what does $\binom{n}{k}$ count?
8. Your turn: add the first $10$ terms of $2, 5, 8, \ldots$, step 2
19. Your turn: how many terms are in $(a + b)^{10}$, and what is the middle coefficient of $(x+y)^6$?, step 2